text:
Does altimeter setting affect the vertical guidance in a LPV approach?
I had assumed during a WAAS approach when the GPS is in LPV mode, both lateral guidance and vertical guidance are generated from GPS/WAAS signals and the pitot-static system should have no effect on its vertical guidance. This is in contrast with LNAV/VNAV approach, where the vertical guidance comes from the barometric system, the same source as altimeter.
However, when I read RNAV (GPS) Z RWY 30 @ KHAF, I will get a different DA if I use a different altimeter setting. So, why does altimeter settings matter to the DA in an LPV approach?

To directly answer your question, "...why does altimeter settings matter to the DA in an LPV approach?", it is because the DA is still predicated on the barometric altitude.

@Porcupine911: It make sense. The minimum (DA in this example) given in the approach plate is baro-minimum, as against radar-altimeter minimum or GPS-altitude minimum.

Short Answer:
No, the vertical component of LPV approach performance is based on WAAS GPS input, not the altimeter or barometric system.
Longer Answer:
Your assumption about LPV performance is correct: Baro-VNAV is not used for LPV approaches.
On any type of approach with a Decision Altitude (DA), the aircraft's arrival at the DA is determined by the pilot's reading of the aircraft's sensitive altimeter.
If the altimeter is set to an altimeter setting report from a distant reporting station, SFO in this case, then the DA may be increased to give additional altitude margin to allow for variance in local atmospheric pressure.
This is true for the DA or Minimum Descent Altitude (MDA) on any approach, whether a GPS approach or not. See the MKT ILS 33 and MKT RNAV 15 for other examples of this. See the following excerpts from the Aeronautical Information Manual on altimeter settings for approaches (emphasis mine):

5-4-5
a.
4. Approach minimums are based on the local
  altimeter setting for that airport, unless annotated
  otherwise; e.g., Oklahoma City/Will Rogers World
  approaches are based on having a Will Rogers World
  altimeter setting. When a different altimeter source is
  required, or more than one source is authorized, it will
  be annotated on the approach chart; e.g., use Sidney
  altimeter setting, if not received, use Scottsbluff
  altimeter setting. Approach minimums may be raised
  when a nonlocal altimeter source is authorized. When
  more than one altimeter source is authorized, and the
  minima are different, they will be shown by separate
  lines in the approach minima box or a note; e.g., use
  Manhattan altimeter setting; when not available use
  Salina altimeter setting and increase all MDAs
  40 feet. When the altimeter must be obtained from a
  source other than air traffic a note will indicate the
  source; e.g., Obtain local altimeter setting on CTAF.
  When the altimeter setting(s) on which the approach
  is based is not available, the approach is not
  authorized. Baro−VNAV must be flown using the
  local altimeter setting only. Where no local altimeter
  is available, the LNAV/VNAV line will still be
  published for use by WAAS receivers with a note that
  Baro−VNAV is not authorized. When a local and at
  least one other altimeter setting source is authorized
  and the local altimeter is not available Baro−VNAV
  is not authorized; however, the LNAV/VNAV
  minima can still be used by WAAS receivers using the
  alternate altimeter setting source.

Note that Baro-VNAV is a type of VNAV system which uses barometric
altimetry as a basis for vertical navigation performance. An LNAV/VNAV approach can be flown using WAAS equipment without a Baro-VNAV system.

So what you're saying is that, although vertical guidance is being provided by WAAS the DA is set by baro altimeter?

@TomMcW The DA is the altitude at which the pilot decides whether to continue with the approach or go missed; the pilot determines arrival at this altitude by reference to the aircraft altimeter. Baro-VNAV is a vertical navigational system to which gives the required navigational performance for the VNAV component of LNAV/VNAV approaches (though other systems such as WAAS can also provide this level of performance). Does that answer your question? I wasn't quite certain what you meant by baro altimeter.

I guess what I'm summarizing from your answer is that the although the WAAS guidance is not affected by the altimeter setting the FAA still sets the DA by the pressure altimeter and not the altitude calculated by the GPS. Is that correct?

@TomMcW That's correct. You probably already understand this, but in an attempt to be perfectly clear, let me reiterate that pilots rely on the indicated altitude as seen on the sensitive altimeter, not the altitude given by the GPS (which may be the true altitude, depending on the GPS precision).

Why is that? Doesn't WAAS GPS give you a more accurate and reliable attitude that is not dependant on QNH?

@TomMcW Yes, WAAS altitude is very precise and reliable. That's a good question—probably a good new question on it's own! The answer is long and includes the recent development of precise GPS and the usefulness of having all aircraft with altimeters set the same. It's simpler.

I figure it's so everybody is on the same page as far as altitude is concerned concerning separation. Same reason everybody uses 29.92 above TA. It just seems for approach and landing purposes, though, you'd want to use whatever is most accurate so you don't bump into terrain.

This answer is dead on. If the missed approach point was a specified distance from the runway instead of a specified altitude, then the altimeter setting wouldn't matter because all guidance would be generated by the GPS signal.  Since you are referring to the altimeter to determine when to go missed though, you could possibly continue following the GPS signal to an altitude lower than intended if the altimeter is not set to the correct altitude for the field.  Maybe that will help to understand why it matters though if you determine the missed approach point by the altitude instead.


text:
Why is the VFR traffic pattern area at some airports different from its IFR circling approach area?
Take Half Moon Bay Airport (KHAF) as an example.

Under VFR, the traffic pattern is right pattern for RWY 30 and left pattern for RWY 12. This implies the downwind leg is to the EAST of the RWY 12-30, and therefore we should fly over the land at TPA 999'.
However, looking at the IFR plates, both RNAV (GPS) Y Runway 30 and RNAV (GPS) Y Runway 12 indicate we are required to circle WEST over the water at 800', since that "Circling NA east of Rwy 12/30".

I understand it is likely that we have more vertical room over the water since there are no obstacles there. However, I don't get the reason to require VFR traffic over the land. Is it for VFR/IFR traffic separation?
Some reasons that I came up with

Favors the water, against the land: Terrain/Obstacles, Noise Abatement.
Favors the land, against the water: Marine Sanctuary Protection.

It looks like you have a few different questions here. As a caveat, I know nothing about Half Moon Bay so these are (educated) guesses to some extent. Someone with personal experience at KHAF could probably give better answers.
Why is circling always west of the runway at KHAF?
Circling areas are planned according to the TERPS criteria (see section 2-7-1); the formulas are in the document, but the basic idea is to make sure that there are no obstructions to circling within a certain distance of the runway. It looks very likely that at KHAF the terrain east of the field rises too sharply to allow safe circling there: the minimum allowed in the TERPS is 1.3nm/1.5sm/2.4km, and Google Maps shows that there's definitely rising terrain within that distance.
Why is the pattern always east of the runway at KHAF?
I don't know for sure. Noise abatement is obviously a concern at KHAF but then as you said, why not put the pattern to the west, over the ocean? The most obvious reason is that Half Moon Bay is in the Monterey Bay National Marine Sanctuary and flying over national parks below 2000' AGL is discouraged. Another possible reason is to avoid directing light, single-engine aircraft to fly over water unnecessarily.
Why - in general - would a certain area be used for the pattern but not for circling?
There's no reason why one area should always be suitable for both needs. In particular, circling requires guaranteed obstacle clearance within a certain distance from the runway. VFR pilots should be able to see and avoid, but obviously that assumption isn't possible for IFR. If you fly a circling approach at minimums then you might even go back into cloud while circling (and then go missed); in that scenario you really don't want to have to worry about terrain or obstacles being anywhere nearby.

On the [map provided by NOAA(http://montereybay.noaa.gov/materials/maps.html#overflight) of the Monterey Marine Sanctuary KHAF is not in the overflight restriction zone. And the noise abatement link you provided shows "extremely noise sensitive ares" within the designated pattern area, not to mention the hospital you're supposed to avoid.

It’s because of the marine sanctuary. 99% of the time it’s Runway 30, and there are noise abatement procedures (pdf) for the east side traffic pattern as well.

Although it's in the sanctuary zone, according to their map it's not in the overflight restriction zone

Just because it's not in the zone that doesn't mean they want traffic over it

The restriction zones are printed on the sectional as magenta lines with dots inside them. Whereas the sanctuary boundaries themselves are not depicted.There are several airports located on the sanctuary but outside the restriction zones. The MRY class C even has a portion over the restriction zone. It looks like the circling area has an MDA of less than 1000 feet. (Let me know if I'm wrong about that. I'm not good at reading app plates) Not saying the sanctuary is a factor, but I don't see any evidence that that is the case

I live in Moss Beach next to KHAF. Specifically, they want the VFR traffic over land because of the Fitzgerald Marine Reserve to the NW of 30. There's a huge pack of seals w/ cubs on the beach, so they'd rather keep the noise over the residential areas. You wouldn't want to send IFR traffic over that land though, not with Montara Mountain to the N and it getting steep pretty quickly to the NE.

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text:
Why depict TRSA on a map?
Participation in TRSA services is completely optional. The depicted airspace is all still class E, and you can fly straight through it sans radio or transponder without violating any regulation.
Obviously, if you do talk to them, they can provide radar services, but this is no different from any other class E airspace owned by an approach facility, including (notably) the class E above and below the depicted TRSA boundaries.
So, why bother showing it at all?  And even if there’s a reason to show it, what’s the point in such complex configurations that have zero impact on either pilots or ATC?

Related-- https://aviation.stackexchange.com/q/22134/34686 (see all answers) .  It seems to me that a) the reason it is complex is that the radar coverage naturally extends further at higher altitudes, and b) the reason it is depicted on the map is because it is complex.  If it were simpler, like a single ring at fixed radius, it might not be depicted, seeing as it has no regulatory impact.

@quietflyer not all TRSAs are excessively complex, e.g. both BPT and LCH are simple concentric rings (though there are more rings than a Charlie would have).

@randomhead -- re "simple concentric rings" -- I guess my concept of "not complex" would be a single ring of set radius which is the same at all TRSA airports, with coverage extending from ground level to a set AGL altitude which is the same at all TRSA airports.  Similar to the old pre-1993 "Airport Traffic Area" concept which defined the airspace volume around a towered airport within which establishment of communications was required before entry-- and which in fact was not depicted on the VFR sectional charts.

I think - and I may be wrong - that your question assumes that ATC services are the same in a TRSA and in class E in general. Or perhaps that TRSA Service and flight following give you exactly the same thing.
In fact, TRSA service is slightly different: it provides everything that Basic Radar Service does (e.g. safety alerts, sequencing), plus separation services for all participating aircraft. Usually, ATC doesn't provide any separation service to VFR traffic in class E.
The ATC Orders (JO 7110.65) sections 7-6 and 7-7 cover Basic Radar Service and Terminal Radar Service; you can read more details there. Since the ATC services available in a TRSA aren't the same as those available in general class E airspace, that seems to me like a valid reason to chart it.

I wasn’t aware of VFR/VFR separation service, which isn’t available in normal class E; that explains it!

@StephenS Yes, it's an odd one. AFAIK, the only other airspace where VFR traffic is separated from all other traffic is class B.

@757toga Right, but they aren't separated from other VFR aircraft. in TRSA they are.

If you visit the Aeronautical Information Manual, section 3-5-6, it will provide the answer for you:
"TRSAs were originally established as part of the Terminal Radar Program at selected airports. TRSAs were never controlled airspace from a regulatory standpoint because the establishment of TRSAs was never subject to the rulemaking process; consequently, TRSAs are not contained in 14 CFR Part 71 nor are there any TRSA operating rules in 14 CFR Part 91. Part of the Airport Radar Service Area (ARSA) program was to eventually replace all TRSAs. However, the ARSA requirements became relatively stringent and it was subsequently decided that TRSAs would have to meet ARSA criteria before they would be converted. TRSAs do not fit into any of the U.S. airspace classes; therefore, they will continue to be non-Part 71 airspace areas where participating pilots can receive additional radar services which have been redefined as TRSA Service."
https://www.faa.gov/air_traffic/publications/atpubs/aim_html/chap3_section_5.html
You do not need to participate, but it is a service which is available to you should you choose to take advantage of it.  When transitioning those areas, you are wise to take advantage of it.

This is (more detail on) what I said in the first paragraph of the Q but doesn’t provide an answer to the question.

The easiest way that I was told to relate to a TRSA is to think of it as a baby Class C airspace. The TRSA is technically Class D or E airspace with almost all of the equivalent facilities and equipment of a Class C. You can receive almost all of the same ATC service of a Class C in a TRSA. But, it is not mandatory to use them.
Another way to view a TRSA is that many of them were formerly, are currently striving to be, or soon will be Class C airspace. Class C airspace. They are just currently downgraded to Class E or D.

Allegedly, AOPA and EAA have lobbied the FAA to not convert any more to class C due to the mode C/ADS-B/radio requirements. They’re permanently stuck in limbo.

As Pondlife's answer points out, it isn't as simple as "baby Charlie." The separation services that a participating VFR pilot receives are more than they would receive in a Charlie... but participation is not mandatory and equipment requirements are not as stringent. I wouldn't say it's "baby" or "grown up." It's just different.

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text:
Pilots with one eye (monocular vision), versus monovision contact lenses
If it is allowed for a pilot to have one eye (monocular vision), why shouldn't it be allowed for a pilot to have monovision contact lenses? With monovision contacts one has more peripheral vision than with one eye, and overlapping vision. Depth perception is not affected that much by monovision contacts.

Does this answer your question? Can you obtain a pilot license with impaired vision / being blind in one eye?

That "dupe" doesn't even begin to address what's asked in this question; I didn't see any reference to monovision contact lenses there.  This question might already have an answer here in Av.SE, but that question most certainly isn't it.

Welcome to aviation.SE! Which country or regulations are you asking about? The rules may be different in different places.

It's probably because monovision contact lenses have a downside in that some people find it compromises the clarity of their distance vision too much, making distant objects appear slightly blurred. This would hinder your ability to see other aircraft, for example.

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text:
When it comes to private airports those marked on sectionals with (Pvt) what frequency's do they use if any at all
As the title says for private airports and planes that fly in and out of them is there a frequency that they use and if so where can it be found?

Referencing AIM:  Services Available to Pilots/Summary of Recommended Communication Procedures (Link).
No Tower, FSS, or UNICOM
-Self‐announce on MULTICOM frequency 122.9
However - as you may expect, the short of it is - it's private - so nobody will be listening most likely.  I'm only listing the AIM reference as the most authoritative source - which even then - likely doesn't even fully cover this.
I'd have to imagine that some of these are owned by individuals living on a ranch like setting in a remote area, and assume that they are going to be the only individuals using the runway (unless an emergency happens).  Even if an emergency happens, the individual is unlikely going to be listening to this radio.  In this "individually owned" scenario, it's also unlikely any pattern work is going on.
In other situations, these may be owned by a flight school or a private group.  I'd have to imagine in this case - they select one of the common CTAF frequencies also published in this table.
General Note:   The owners of these opt to chart them for your safety in case of an emergency, and not much else.  Some may be more generous and allow you to reach out to them with advance planning in certain limited circumstances.  The bottom line is to realize that these fields are private for a reason.

FAR's specify that private airports can not have a CTAF frequency assigned. The most correct answer is MULTICOM 122.9 as mentioned above, per AIM section 4-1-11, table 4-1-2. As far as nothing much happening at private airports, I've been flying from one for over 15 years and we routinely have 10-15 aircraft based there. Yeah, there is radio traffic. Now, since you're not supposed to land at a private airport without permission, you should just ask what frequency when you ask for permission to land, right? There are a few private airports that still use 122.75 (air-to-air) which at one time was also for them too, but the FAA dropped that frequency assignment from the FAR/AIM a few years ago.

what FAR talks about that?


text:
Distance where transition from laminar to turbulent boundary takes place
How do we work out where the distance from the leading edge the laminar boundary layer would turn into a turbulent boundary layer? Assuming we have been given the length of the plate, speed, dynamic viscosity of air and the Reynolds number?

For help with homework problems, always show what you have tried so far.  Otherwise you come across as asking people to do your homework for you, which I assume is not what you wish.

Related: Is airflow ever laminar in the boundary layer, near the wing leading edge?


text:
Relationship between hover power and climbing power
I've been told that, generally speaking, rotor efficiency increase with rotor diameter. This is because the thrust generated by a helicopter rotor equals the mass of air moved times the delta V of the air moved, while the kinetic energy imparted to the air by the helicopter is proportional to the mass of the air times the square of the delta V. Therefore, since increasing the rotor area increases the mass of air moved, and since hovering requires a constant thrust force, increasing the rotor area will decrease the delta V of the air. And according to E = 1/2 m*v^2, doubling the mass and halving the delta V of the air will decrease the energy imparted to it.
From this, you can theorize that a helicopter with an infinitely large rotor diameter would require zero energy to remain hovering.
Now, assuming this is all correct, let's assume that the helicopter briefly accelerates upwards until it is moving vertically at 1 mph. Let's also assume that wind resistance is negligible. As it moves steadily upward, it will need the same force as when it was just hovering. And the above analysis suggests that the helicopter with the infinitely large rotor would not need to consume any energy to maintain the constant upward speed. But we know that a helicopter climbing into the sky is gaining potential energy, meaning that this conclusion of zero work done by the helicopter can't be correct.
So, what am I missing? Why does the analysis that suggests that neither a hovering helicopter nor a steadily rising helicopter with an infinitely large rotor need to consume any energy?

So on your theoretical helicopter with the infinitely large rotor, is the rotor stationary? If so, how does it impart any downward thrust to allow it to hover? And if not, what causes it to rotate?

The rotor is powered by an engine. To achieve steady thrust, the rotor needs to spin at some speed and will consume some amount of power. As the rotor increases in size, the amount of power needed to stay aloft decrease. In the limit where the rotor diameter approaches to infinity, the power requirement approaches zero.

So we have an infinitely long rotor, hovering over an infinitely long/wide flat earth, powered by an engine that had infinite power to get those infinitely long blades moving in the first place, but perhaps no power required to keep them moving, so that all the air in the universe is being swept by these infinitely long blades, whose tip velocity would be infinite except that there ARE no rotor tips (tho much of the blade itself will be supersonic+ )... we're pretty far into the realm of theoretical / detached from reality here. VTC as not being about aviation, as defined in the Help Center.

@RalphJ Really? Helicopters and rotors are not about aviation? Voting to Leave Open, welcoming PrestonBecker to this site. Very good question.

Infinitely long rotor blades have one problem: they cannot be rotated, as they would instantly exceed the speed of light at their infinitely far away blade tips. One should not use infinities in calculations such as this.

@Jpe61 An infinite rotor creating finite lift will turn infinitely slowly. Don't look at the infinite case but several with increasingly large rotors and see where this is heading. The asymptotic case will be the solution.

You see where this is going? Nowhere. No point in playing with infinities in cases like this.

Exactly what Jpe61 said - extrapolating a simple equation to infinity serves no practical purpose, and ignoring physical boundary conditions moves the discussion rapidly from a potentially educational abstraction to complete absurdity.  (i.e. does "infinitely slowly" equal stopped?!)  I'm DVing the question for gross violations of common sense.  (i.e. "not useful"...)

"As it moves steadily upward, it will need the same force as when it was just hovering." WRONG!  It will need more force -> energy input to overcome gravity.  If it didn't, you could use it to create a perpetual motion machine.

@Koyovis, Yes... really! I agree with Ralph J:  Musings about the function of helicopter blades that exceed the diameter of our planetary atmosphere, no, the solar system, no, the known universe and beyond have no bearing whatsoever on "aviation" as we experience it in actual reality. (i.e. defined as the finite physical interaction of an air vehicle within the confines of earth's atmosphere and gravitation field...) I'm not voting to close because I think debunking absurd conclusions can be useful, but I agree that this is a hypothetical physics question with no practical application.

@MichaelHall Aviation is ruled by physics, and Einstein has shown us how useful thought experiments can be.

@Koyovis, to the extent that thought experiments are bounded by reality, I agree.

@MichaelHall I didn't realize how much trouble I would stir up by invoking infinity. Perhaps I should have talked about the rotor blades getting arbitrarily large, or I should have talked about the asymptotic values for energy. With that said, talking about infinity was quick and dirty, and in the end led to the excellent and insightful answer by Peter below. And without already knowing about inflow angle, I'm not sure if there was a different way to ask the question to get the same answer. So, sorry to ruffle feathers, but as they say, all's well that ends well

No worries, it's all just chit chat.  No ruffled feathers here!  Cheers.

We don't need to look at the infinite case to see where the error in your thought experiment is. It will be sufficient to see what is different between a hovering rotor and an ascending one.
Basically, the angle of inflow into the rotor is different between the two and this will not change as diameter is changed. Without considering downwash and induced effects (which are zero in the infinite case), the inflow angle of the hovering rotor is zero, resulting in a lift vector which is exactly perpendicular in case of the infinite rotor in inviscid flow. This means lift is pointing straight up. Its creation does not use engine torque, as you correctly observed.
Now consider the ascending rotor: Here the inflow angle is somewhat positive, namely the angle defined by the inverse tangent of climb speed divided by rotational speed at the radius concerned. This inclined inflow angle will tilt the lift vector backwards (against the direction of rotation) such that the horizontal component of lift will cause a moment which acts against the direction of rotation. This moment needs to be overcome by added engine torque. And you need to adjust collective pitch in order to trim the helicopter for this climb.
Change the climb into a descent and you will see that the same mechanism will add rotational energy to your rotor. Now the inflow angle is from below, tilting the lift vector into the direction of rotation. This is how gyrocopters work!
If you make the rotor larger, the rotor extension will see a very shallow inflow angle because most of the angle change resulting from a climb speed occurs at the inner rotor. Thus, going to infinity will not change the picture.

Regarding a hovering rotor requiring 0 torque (end of 2nd paragraph above)... is there any distinction between such a rotor spinning on a helo sitting on the ground, and the rotor spinning on (i.e. supporting) a helo hovering a few feet off the ground?

@RalphJ In real life yes because of ground effect. In this case with an infinite rotor no since downwash is infinitely small.

@PeterKämpf Thanks for this answer. It is exactly what I was looking for.
Where did you learn about inflow angle and everything else you discussed? Was if from school or is there a good book or webpage out there on the subject? I have a Mechanical engineering background myself and, while not having not done a lot of work with aerodynamics so far, am hoping to learn more about the topic. (not that I am trying to give you extra work, but if you have any sources off the top of your head, that would be cool)

@PrestonBecker Oh boy, I am a voracious reader but cannot point to a single source. Things that interest me simply get stuck in my mind on first encounter. All that nonsense about sports or celebrities, on the other hand, is soon forgotten. At the root is the desire to drill down to the exact explanation for what is around us. And decades of reading and questioning the correctness of what I have read. Remember, especially on the Interwebs you will find a lot of silly and unscientific stuff. That might even include some of my answers here.

The engine must always produce the kinetic energy of the accelerated air mass = ½$\dot m \cdot \Delta V^2$. Equals engine torque times rotational velocity. Engine torque cannot be zero.

@Koyovis For an infinitely large rotor creating a finite lift, $\Delta$v and acceleration tend towards zero.

@Koyovis The thrust equation is T = mv, as you noted. Solving for v gives v = T/m. Substituting that into E = 1/2mv^2 gives E = 1/2m*(T/m)^2 = E = 1/2 T^2/m. As m tends approaches infinity, then the energy term approaches 0. Also, how are you getting the delta symbol and the dot over the M? Is that all through Markdown?

@PeterKämpf That is too bad. I guess I'll have to start scanning the web with "inflow  angle" as a search term

@PrestonBecker through markdown indeed. The equation is put between $ signs, the dot via \dot m.

In a loss free world, many perpetual mobile could exist. But unfortunately in the real world, we can be very certain that there never is such a thing as a free movement lunch. A hovering helicopter has the following losses:

Profile drag and induced drag, both quantified in this answer. An infinitely long rotor blade has zero induced drag, but unfortunately an infinite amount of profile drag.
Parasitic drag. From rotor/body interference, and from the vertical velocity when climbing.
Power loss from the tail rotor, or interference loss from the coax main rotors.

So in the hover, engine power is required for compensating losses in creating thrust = $ m \cdot \Delta V $. With $\dot m$ being the mass flow of air through the rotor in [kg/s] and $\Delta V $ the increase in velocity of the airflow through the rotor. So the equation is not dimensioned in W but in N, not in energy but in a static force. So indeed the whole thought experiment with infinitely long rotor blades is unnecessary. Whatever the length of the blades is, the engine power is necessary to overcome losses in creating hover thrust.

But we know that a helicopter climbing into the sky is gaining potential energy, meaning that this conclusion of zero work done by the helicopter can't be correct.

In a climb, the thrust is higher in order to overcome the vertical drag component. The extra energy required for the climb is equal to $F \cdot c$, with $F $ = thrust and $c$ = climb velocity. As everyone instinctively understands when they can hear their car engine revving up when running up a hill in cruise control.

True perpetual motion machines not only keep going forever, they allow the builder to extract energy: https://en.wikipedia.org/wiki/Perpetual_motion

At hover an infinitely large rotor would move an infinite amount of air downwards at zero speed.  If the helicopter is ascending though, it won’t see a zero vertical airspeed and so will need to input some energy in order to balance the increasing potential energy.

Rotating infinitely long rotors takes an infinite amount of energy.

And an infinite amount of material, and you’d need an infinite atmosphere to fly in.  This is, I hazard, largely a thought experiment.  Nevertheless in this fanciful scenario the amount of energy that would need to be expended in supporting the aircraft would be zero.

In order to hold a mass m at a fixed height h within a gravitational field of acceleration g, you don't need any energy. A pillar is enough...
But in order to increase that height by ∆h, you have to add energy. Exactly, m·g·∆h

You have fallen into a common mathematical trap.
You note that "increasing the rotor area increases the mass of air moved." So an infinite rotor moves an infinite mass of air. Mathematically, 1/infinity = 0, or infinity = 1/0.
So your energy equation E = 1/2 m*v^2 becomes:
E = 1/2 (1/0)*0^2 = 1/2 0/0
But 0/0 is undefined (as is infinity*0), so E is undefined.
Thus, infinity is what is called a boundary condition, at which the mathematical model breaks down. This is in accord with reality, since no infinity has ever been observed in nature.

"This is in accord with reality..."  Reality, the ultimate boundary condition!  ;)

@Guy Inchbald
Well, since we are dealing with limits, we can use L'Hôpital's rule to determine what the value of the term is
L'Hôpital's rule says
lim(x->c) f(x)/g(x) = lim(x->c) f'(x)/g'(x)
Therefore
lim(x->0) 1/2 * (x^2)/x = lim(x->0) 1/2 * (2x)/1 = 0
Therefore, E = 1/2 1/x * x^2 as x goes to 0 gives that E = 0

@PrestonBecker You appear to have simply substituted x it for 0 and expected your equation to be valid. That is not how L'Hôpital's rule is applied.

The accelerated air mass is infinitely large, the velocity approaches zero. There is no divide-by-zero in the equation.

@Koyovis That is quite beside the point. As my answer states, mathematically, 1/infinity = 0, or infinity = 1/0. Solving equations always involves such manipulations between equivalences, it is how maths works.

It is exactly the point. There is no division by zero in your equation. Stating that infinity is equivalent to divide by zero is incorrect, divide by zero is undefined. Even the limit of (x -> 0) 1/x is undefined, since it can be plus or minus infinity.

@Koyovis These issues were all laid to rest long ago. There is not space in a comment for a fully rigorous proof, so one can only give the gist of it. If you are suggesting the OP's arguments stand up, you are hopelessly off the mark. Feel free to disagree.

In the original question, I spoke about "plugging in" infinity for the mass flow term. That was imprecise. I should phrased the question about evaluating the limit of the energy equation as the mass term approaches infinity. Solving the thrust term for velocity gives V = T/m. And then plugging that into the energy equation gives E = 1/2 m * (T/m)^2. Finding the limit of that equation as m goes to infinity seems like a textbook application of L'Hôpital's rule. Also, the equation simplifies to E = 1/2*T^2/m, which you should be able to analyze as m goes to infinity. Or am I missing something?

L'Hôpital's equation sets you to finding the ratio of two differentials. T is constant in the hover, so dT/dt=0 whatever the value of m or E or anything else. That alone should warn you that your ship is sinking. Feel free to disagree, I am done with this nonsense.

-1 for the statement “Mathematically, 1/infinity = 0, or infinity = 1/0.”

@Koyovis Are you guys a tag team? Like I said, the statement is not rigorous but it gets the gist of the OP's mistake across. This is the Aviation SE, not the Maths SE.


text:
What is the principle behind flight of airplanes?
What is the principle behind flight of airplanes? Need a simple answer for grade 7 student. Thank you

Stick your hand out the window of a car and...

Obxkcd: "airfoil".

For Seventh Grade?
How about:
Because air consists of tiny little particles called atoms, or molecules, and when an airplane moves through the air, it collides with these particles and they bounce off the airplane and push on it, just like if you hit, or push on something with your hand.

This is actually quite a good description for how lift works in the supersonic regime. Subsonic, not so much.

This is exactly how lift works at any speed. If it's not impact of air molecules bouncing off the surface of the aircraft, then what is the mechanism of force transfer? Lift is a force. Where does the force come from, if not from the impact of the air hitting the aluminum.?

@CharlesBretana Of course the air molecules will hit the wing, but they also interact with each other. That interaction is crucial for subsonic airflow around the wing. Neglecting it is an incorrect over-simplification. Therefore, explaining lift on the molecular level is not going to work.

This answer is actually quite brilliant, hard to make it any simplier...

@Bianfable, Of course air molecules hit each other. They hit each other all over the planet, but the only collisions that can exert a FORCE on the airframe are the collisions between air molecules and the airframe. Collisions between air molecules running into one another only exert forces on each other, not on the airframe.

And air molecules are colliding with one another all over the place, all the time, both near an airframe, (within the boundary layer) and further away from the airframe (outside the boundary layer), regardless of whether the airframe is moving through the air above or below the speed of sound.

Perhaps in my earlier comment I should have said "This  is how all aerodynamic forces, including lift, drag, propeller thrust, etc.) work.  Lift is just the component of all the vector sum of all the forces exerted by the molecules hitting the airframe that lies normal (perpendicular) to the aircraft velocity vector.  Just as drag is that component which is parallel to the velocity vector.

@CharlesBretana Unfortunately, your description of lift on a molecular level is still incomplete (what about the upper surface of the wing?). Please read this excellent answer by Peter Kämpf to get a better understanding of how lift actually works.

@Bianfable, Unfortunately, you appear to not understand my answer. First of all, what makes you think I am only talking about lift? ALL aerodynamic forces are the result of molecular collisions. Secondly, what makes you think I am saying anything different about the upper surface of the wing? Air molecules are colliding with the surface of the entire airframe, Top & bottom of the wing, leading and trailing edges, canopy, tip of propeller spinnor, and the tip of the tail-cone, every single point on the surface. I fully understand how lift works.

I have an Aeronautical Engineering Master's degree. I distinctly remember, from one of my courses, having to derive the Ideal Gas Law (PV=nRT) from basic Physics (Newton's laws of motion), and statistical analysis of the multiple collisions of gas molecules on the walls of a container. The point here is that ALL aerodynamic forces are the direct result of these collisions.

Every other "explanation" for the artificial Psuedo forces we call Lift, Drag, etc. etc. are engineering simplifications. Just because you remember a picture of an airplane with a giant Lift Vector sticking out of it pointing up, and another one labeled Drag pointing backwards, doesn't mean that these forces actually exist. They are representations of a Integral Calculus process that sums all the forces from the trillions of molecular collisions that actually are pushing on the airframe.

A basic principle of Physics (Mechanics, actually) is that no matter how many different forces are applied to a body, or what direction they are all applied, nor what point on the body they are applied at, they can be equivalently represented by one force acting through the center of mass, and one rotational torque (or moment) acting on the body on the appropriate rotational axis.  Lift and drag are just the components of this vector summation of all the individual forces from the molecular collisions that are normal to (Lift), and parallel with (Drag) the velocity vector.

Downvoted because this explanation, while technically correct, doesn't really explain anything.

@nick012000,  This is for a seventh-grader!  Would you be trying to make an aeronautical engineer out of him/her? The aggregate patterns that only appear when you examine things in detail do not explain the basic causal forces. The basic forces come about from exactly what I describe.

We forget how esoteric and confusing our analysis of these patterns can be. Lift is caused by 1) the air is moving faster on the top... 2) because the wing pushes the air downwards ... 3) because the wing exerts a down force on the air .... , etc. etc. etc. None of these explains what the actual force is, and are, although technically true, they do not explain the FORCE on the airframe. The collisions of the molecules concept does. All these other explanations do is explain patterns of higher level aggregate properties.

@CharlesBretana The higher-level aggregate properties are what are important to understand, however.

Yes, sir, I agree. I could not agree more - for Engineers.. Not for a seventh grader.  And, honestly, even engineers need to remember at times what the underlying physical principles are that these high level aggregate properties spring from...  lest they start to believe that the high level aggregate properties are in reality the bottom layer of the conceptual hierarchy.

And to quibble even more, the collisions between molecules and the metal (or fabric, whatever) of the airframe, well, that's not the bottom layer either. Because the molecules don't really come into contact with the airframe, only their electromagnetic fields interact. And who knows what other layers lie beneath that that we haven't even discovered yet.

Airplanes fly mainly because their wings push air downwards. As the air is being pushed down, an upwards force pushes the wing up.
The downwards movement of the air is achieved by two methods, both of which are used on airplanes:

The leading edge of the wing is higher than the trailing edge, making the wing act as a wedge. The angle between the line from leading edge (front edge of the wing) to the trailing edge (rear edge of the wing) and the airflow coming towards the wing is called the angle of attack.
The wings profile is such, that it forces the upper airflow go faster than the airflow under the wing. When these airflows meet at the trailing edge, the faster airflow coming from the top tilts the airflow down.

A portion of the lift comes from the faster airflow on top of the wing: as the airflow accelerates, the pressure of the air goes down, causing suction on the top of the wing.
To achieve lift, the airplane needs to go forward, so it either needs an engine to push or pull it, or it needs to glide downwards.
To get more lift, airplanes either raise the angle of attack (raising the leading edge even more higher than the trailing edge), or they go faster. To get less lift, the do the opposite. Adjusting lift will, of course, make the plane go up or down.
If the angle of attack gets too high, the air can no longer follow the curve of the wings upper surface, and the wing will lose its lift. This is called a stall. A wing stalls if the aircraft is going too slow, therefore all aircraft have a minimum safe speed.
Obviously, there is also a maximum speed for airplanes, usually limited by the airplanes structure which can no longer stand the stresses the fast moving air causes.

"When these airflows meet at the trailing edge" -- This sound suspiciously like the "equal transit time" theory, which is incorrect (and imo takes the student further from understanding the idea). On reading that point a second time, I don't think it's what you meant, but you may want to clarify.

A humble suggestion: When explaining to a 7th grader, it helps to use simpler terms like "front of the wing" instead of "leading edge" etc. If I was in 7th grade, I'd also want to know why the air at the top moves faster: https://physics.stackexchange.com/questions/13030/why-does-the-air-flow-faster-over-the-top-of-an-airfoil

There are no two methods, just one. The leading edge being higher than the trailing edge is essential, but the air flowing over the wing being faster is just effect of the pressure decreasing as the air is made to follow the downward sloping surface. It's not a separate contribution, it's the mechanism how the force is transferred to the wing.

Yshavit & Jan I'm trying to keep this as simple as possible, while not drifting too far from the truth. By no means do I subscribe to the equal transit theory, but referring that would have complicated the answer. I think the wing profile is very essential, thats open for dicussion but maybe not here. Nav I thought terms leading edge and trailing edge were self explanatory, but I think your suggestion is better than my original guess. Edited.

Hmmmm... It would be nice to know why my answer was wrong. I mean I really don't need the extra two points the downvote takes, but really, what is so wrong here?

As Jpe61 wrote, the important thing is that the air gets pushed downwards. Contrary to what's often said, it doesn't really have anything to do with the Bernoulli effect.
Flow on the underside of the wing is quite easy to understand: air collides with the surface, which it can't penetrate, so it is forced to change its path from moving horizontally, to moving diagonally down along the wing.

At the point where the path bends, a downwards force needs to act on the air molecules. By Newton's third law, the opposite force must act on the wing, and that's what pushes the plane up.
Easy, right?
...Well, that's actually not sufficient. Consider a wing shape like the following, where you also have the downwards path-deflection, but still don't get any net lift:

You can see there's no lift from the fact that the air flow after the wing is again horizontal, i.e. no vertical momentum has been imparted on the air.
Note in particular that two commonly mentioned factors are not sufficient to generate lift: 1. front edge of the wing higher than aft edge 2. upper surface longer cord than lower surface.
So the crucial thing is that you manage to keep the air streaming diagonally downwards after passing the wing. That's where the importance of the upper surface comes in: it must form a smooth enough curve so the air will follow it, streaming tangentially. In this case, the streams above- and below the wing will both be moving downwards at the trailing edge, so they can just merge again and continue diagonally.

But why would the air even bother curving down above the wing? After all, there's nothing in the way there that would prevent it from moving on in a straight line.
The reason is ambient air pressure. If the molecules did move in a straight line, there would be a whole region of vacuum above the wing. In the boundary between this vacuum and the above air stream, there would be a huge density- and pressure gradient, and that would push air molecules down into the vacuum region. This is a very violent process; it happens when space capsules re-enter the atmosphere at hypersonic speeds, but not in airplane wings.
Instead, with a proper wing profile at appropriate speed and AOA, you get the smooth flow over the wing, which involves no vacuum. It does however involve a lowered pressure right over the curved part of the wing. It is the pressure gradient between this low-pressure region and the higher-pressure ambient air above that maintains the curved flow. And the low-pressure region adds lift too, in addition to the excess-pressure lift from the lower surface.
It also causes a suction effect: higher-pressure air from in front of the wing is accelerated as it gets sucked into the low-pressure region. So the air flow is faster there – Bernoulli's effect. This is, however, not really relevant to the principle how lift works.
At too high AOA, or too low airspeed, or insufficient wing camber, something different happens though: the flow stalls, and instead of moving along the upper surface you get uncontrolled turbulence. This creates strong pressure fluctuations, which also disrupt the flow from under the wing behind the trailing edge. Ultimately the whole flow after the wing is then turbulent and without the downwards momentum, that's why a stalled wing produces no lift or drastically reduced lift.

+1 This is the only sensible answer around that simplifies the more advanced explanation, but does not diverge from it.

I would edit the second picture a bit: the airflow over the wing is faster than below the wing, so the arrowhead should be further out the back. As for the second pic, dunno, maybe the same... Allthough this might step in the region of "too much".

@Jpe61 the arrowhead position has nothing to do with how fast the airflow is.

Oh but of course it doesn't, but it could.

In short, airplanes fly because the shape of their wings and the attitude they move through the air at generate lift.
While there still remains some debate as to exactly how wings do this, science has resolved the reason to a basic principle cause resulting from two different phenomena.
Wings generate lift by altering the momentum of air passing around them.  They do this by the curvature, or camber, of their upper and lower surfaces, which accomplish two things.  First, airflow which strikes the lower surface of the wing is deflected downward, resulting in a change of momentum and a resultant lift force through the conservation of momentum and Newton’s Third Law of Motion.  Think about a child sticking their hand out of a car window with the palm down and fingers pointed toward the front of the car.  As the car drives down the road at speed, the child experiences a force on their hand trying to push it upwards and backwards this is a result of air striking the child’s hand on the palm and being deflected downward, just as air striking the bottom of a wing does the same thing.
The second action is that air passing over the top of a wing is accelerated, which causes a drop in pressure over the top surface of the wing due to Bernoulli’s Principle.  This low pressure area draws down high pressure air surrounding it altering its momentum and resulting in a lifting force.  These two lifting forces constitute the overall force of lift, which counters the force of gravity.  As a consequence the airplane can now fly.
The amount of lifting force that we can generate is directly proportional to the square of the airspeed, the density of the air, and the angle of attack, which is the angle between the chord line and the direction of the airflow to the wing.

Bernoulli's principle explains the air over the wing the other way around. There is a separate argument why the pressure is decreased (due to inertia and receding surface), and the increase in speed is just due to Bernoulli's principle—which makes it tangential to the lift generation, because you've already lift by the lower pressure and don't need the speed for anything.

I don’t know if there’s one way or the other.  Bernoulli‘s principle is the concept of energy conservation, so in the absence of any other factors aside from velocity and pressure, energy is conserved and will necessitate a change in one if the other is altered.

You claim that “air passing over the top of a wing is accelerated, which causes a drop in pressure”, but in fact Bernoulli's principle tells us that the air is accelerated while we already know about the pressure drop! So your claim is not part of the explanation.


text:
How to read the pressure diagram of an airfoil?
According to most authors, "lift is the result of the difference of pressure on airfoil sides".
My question is not about how this difference is generated, but only about the representation of the pressure around the airfoil. In particular, diagrams like this are frequently seen when talking about airfoils:

From Bernoulli or Newton's Laws for Lift?
Being not an aerodynamics engineer, I'll like to understand what can be deduced from this diagram.

Does this diagram indeed shows the variation of pressure on the contour of the airfoil ? Will this vary if the angle of attack or the speed are changed?
Is it correct to deduce from this diagram that the points on the red portion undergo a pressure smaller than free airstream static pressure, and those on the blue portion undergo a pressure greater than free airstream static pressure?

What is the correct method to obtain the relative pressure at a given point p1? Is it to measure the length of the segment which is perpendicular to the airfoil surface at point p1? What is the scale?

On the figure above, is p2 absolute pressure indeed equal to free stream air?
Do point p2 and the similar one found on the trailing edge have specific names?
What other elements can be deduced from the graphic?

Related: How to plot the pressure distribution over an airfoil?

Good questions for a fair understanding of the topic.

for 1. yes and yes
for 5., from my answer on the linked question, they are called "stagnation points"

@Federico Isn't the $C_p$ at a stagnation point 1? p2 seems to have a $C_p=0$.

A nice set of questions indeed, and a set of good answers. Although aren't sets of questions frowned upon here, or are these frowns only for low rep users?

@Koyovis:There a main question: "How to read the pressure at a given point from the diagram" and the other are directly linked to this one. It's like asking about languages spoken by ATCO, and then asking if an operator in China must demonstrate their skills in English.

The drawing in your answer shows the contour of the tips of the local pressure vectors when they are plotted perpendicularly to the local airfoil contour. Yes, this will change with angle of attack. In potential flow theory, the local pressure can be calculated as the linear superposition of a contribution from camber and one from angle of attack. While the camber-related part is constant, the angle-of-attack related part varies linearly with this parameter.
Yes, since the diagram shows the pressure distribution at high angle of attack, the top surface shows mostly suction (= local pressure is smaller than ambient pressure) and all of the bottom surface pressure that is higher than ambient pressure. Static pressure is equivalent to potential energy and total energy is constant, so this means that flow speed is higher than the free flow speed over the red region and lower over the blue region.
Yes, measuring the perpendicular distance between the airfoil surface and the contour line will give you a pressure difference to the ambient pressure, but there is no scale, so all you can do is read relative pressures from that drawing. I recommend to use a different way of plotting pressure, see here for more.
Yes.
P2 has no specific name; here the flow accelerates through the free stream speed. The stagnation point is below the highest distance between airfoil and pressure contour near the nose (in the blue region). That the local pressure at the trailing edge is equal to free stream pressure is a coincidence - at lower angles of attack, the flow will decelerate below the free stream value near the trailing edge, so the blue region would wrap around the trailing edge. Here, flow separation at the trailing edge is imminent, and if the angle of attack is increased a bit, the pressure will turn to suction at the trailing edge.
I think this graph is not quite realistic but is meant to show the general tendency only. More can be seen when inviscid and viscous pressure are plotted together over chord in a cartesian coordinate system.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Is it possible to disengage only one axis of a two-axis autopilot?
Take GFC700 for example, when autopilot (or flight director) engaged, the lateral mode falls into

ROL, HDG, GPS, VOR, VAPP, LOC, BC, ...

and the vertical model has to fall into

PIT, ALT, FLC, VS, VPATH, GS, GP, ...

When disengaged (or temporarily by CWS), a pilot has to hand-fly both axes.
I am wondering if this is universal across all autopilot systems. Is there any autopilot that I may engage one axis and hand-flying the other. e.g. The autopilot only holds altitude and I may roll the yoke without any counter pressure. Are there regulations or purely from technical consideration?

The autopilot (Genesys System 50) in my C-177 can altitude hold while allowing the pilot to control the roll axis. Each axis can be engaged independent of each other.

Some autopilots found on advanced aircraft can be engaged on selected axis only, for example Boeing 737, 747-200 & MD-11. There are three autopilot indicators on the Boeing autopilots: auto throttle, roll and pitch. Each can be in a different mode.
Here a Boeing 737's autopilot roll mode is set to Heading Select (HDG SEL) while the Pitch is flown manually (CWS P):

The pilot can pull or push the yoke, and the autopilot will roll the aircraft automatically. If the pilot steer the yoke left or right, the roll mode will change to CWS R.
In this image below, the pilot is manually Rolling the aircraft (CWS R), but the autopilot pitch mode is set to Vertical Speed mode.

Later models, such as the Boeing 777, does not have this feature - both pitch and roll modes must be engaged together.

@MikeSowsun not really. All Boeing 737 models can. 747-100, 747-200, 767-200 can (not sure about 757). 747-400 & 777 cannot.

You are correct. I forgot about Control Wheel Steering and I completely forgot that you can separate roll and pitch if the aircraft has CWS. From my experience, I seem to recall some 767-200's had it but no one ever used it. None of the 767-300's and 777's have CWS.

On some autopilots it is clearly possible to disconnect each axis separately as evidenced by the accident of Aeroflot flight 593. On others, including the ones used on later Airbus models (A320+; the accident was A310), it is not.


text:
What are the reasons for autopilot restrictions on instrument approaches?
On some instrument approaches, the autopilot-coupled approach is not authorized (or not authorized below a certain minimum).
For example,

Autopilot-coupled NA (MRY: ILS OR LOC RWY 10R)

Autopilot-coupled NA below a minimum (LVK: ILS RWY 25R)

Moreover, on some approaches, the DA is lower if we have a AP/FD/HUD.
I guess the lower DA is because we can give more attention outside the window when there is AP/FD/HUD's assistance. I didn't come up with any idea or context why autopilot-coupling won't work.
If it is turbulent and the autopilot doesn't respond quickly enough to keep the needle centre, we may not use autopilot at all. If the air is calm, the autopilot may well outperform hand-flying. However, it depends on the quality of the autopilot.

It's the fact the autopilot works more accurately than a pilot which is actually the cause of the restriction.
The decision to restrict the use of the autopilot usually comes from the certification agency after the ILS inspection flight detected glideslope erratic variations or reversals. The false signals are likely due to interference from the environment or the nearby activities with the runway equipment.
Flight inspection volume
The ILS volume surveyed by certification agencies using ICAO recommendations is located below a 5.25° angle. It means G/S indications are checked and expected to be correct only below this angle.

Source
This is why G/S are usually intercepted from below unless specific procedures are in place and extra care is given to validate G/S indications.
Anomalies in flight inspection volume
However there can be false and reversed guidance areas within this volume. They are discovered when the ILS is commissioned and flight tested by the certification agency. Example of glideslope reversal during an inspection on the nominal slope:

(Source, page 15-97)
In these areas, the ILS signal may vary quickly and if the autopilot were active it could react to the variations, while a pilot will likely not notice the interference, or will disregard it unless it follows a trend for a few seconds. A pilot with a Vertical Situation Display can confirm the ILS error visually.
A reversal, which is a guidance given in the wrong direction, can lead to a serious incident or an accident, due to autopilot quick reaction time. When such reversals are discovered during the inspection flight, autopilot approaches are not authorized, or forbidden during the final segment where the consequences can be the worst.
Consequences of a reversal during autopilot approach
Using the erroneous signal from the ILS may divert the aircraft from the approach path. Being below the glideslope close to the threshold is indeed not desirable, especially for a jet aircraft for which the thrust increase required to recover the lost altitude cannot be obtained instantaneously.
Suddenly jumping higher is also dangerous. It leads to a loss of airspeed, and possibly to an unrecoverable stall. Below is an example of autopilot reaction to a reversal. While it didn't happen in the certified volume of the ILS, the principle is exactly the one found in any reversal.
Some explanation about the context it happened
It was believed for a long time ILS radio signals for the glideslope contain false guidance at angles multiple of the nominal approach slope (e.g. 6° and 9° for a 3° G/S), but it was also taken for certain these false glideslopes were not reversals and would anyway guide the aircraft to the threshold, at a different descent rate (which would be enough to warn the crew about them following a wrong glide angle).
However G/S reversals, especially after 9° has been observed later (2011-2013), for some types of ILS antennas, without a full understanding of the reasons. Among these cases (source ICAO):

Schiphol Airport, The Netherlands, 2011, KLM, Embraer E190
Murcia Airport, Spain, 2011, Ryanair, Boeing 737-800
Charles de Gaulle Airport, France, 2012, Air France, A340 (BEA report)
Treviso Airport, Italy, 2013, Ryanair, Boeing 737-800
Eindhoven Airport, Netherlands, 2013, Ryanair, Boeing 737-800.

Eindhoven serious incident, May 2013
A B738 on autopilot was flying a non stabilized approach where the G/S had been captured from above. When reaching the 9° glideslope area, at 0.85 NM from the threshold, the aircraft suddenly pitched up to 24° and climbed from 400 to 1,000 ft. Airspeed plummeted to 100 kt, the stick shaker activated:

Source. Original source: Dutch Safety Board report
From the investigation done by the Dutch agency, it appears the actual guidance provided around 6° and 9° is completely reversed:

Source
As a partial mitigation measure, Boeing changed the Flight Control Computer software to limit the pitch up rate in G/S mode. The BEA also recommended manufacturers would prevent the capture of the G/S by an aircraft not on the published glidepath.

I expect the autopilot will be enable to filter out the erroneous noise and take the average. I think the AHRS does the signal noise reduction.

@skyoasis: The AHRS has no ILS input.

I mean since AHRS can a lot of variation filtering due to engine vibration, turbulence to make heading, pitch reading stable. In analogous to AHRS, the NAV receivers should do the same.

@skyoasis: Probably to some degree (albeit it should be confirmed by reading the regulation for ILS receivers), however this doesn't prevent incidents to occur, like at O'Hare in July 2012. A pilot would look at the PAPI if one is available, or to the vertical path of the navigational display, if the aircraft is fitted with one, and detect something inconsistent with the ILS path.

@mins can you provide some authoritative evidence for this answer.

@rbp: I think it's in the Flight Inspection Handbook TI 8200.52, page 15-96 (Rate of Change/ Reversal in the Slope of the Glidepath): "The use of facilities which do not meet the change/reversal tolerance must be limited by a NOTAM (see Chapter 5, Paragraph 5.12) that withholds authorization for autopilot coupled approaches"


text:
Are we expected to fly distanced leg (instead of timed leg) in holding pattern in GPS approaches?
The standard inbound/outbound holding leg are 1 min leg each (no wind condition). I noticed that a distance (4 nm in the following example) is noted in many GPS approaches. Do we need to fly the leg as 4 nm, or the number conveys other meanings?
E.g. http://flightaware.com/resources/airport/TCY/IAP/RNAV+(GPS)+RWY+12/pdf

Short Answer:
Yes, the pilot is expected to fly the hold as depicted, with an outbound leg based on distance (see AIM 5-3-7, paragraph b). Most holds depicted on GPS instrument approaches in the US now have a specified distance for the outbound hold leg, instead of a timed leg.
Longer Answer:
The traditional hold leg distance is determined by timing. See AIM 5-3-7 § j. 4.:

4. Timing.
(a) Inbound Leg.
(1) At or below 14,000 feet MSL: 1 minute.
(2) Above 14,000 feet MSL: 11/2 minutes.
NOTE- The initial outbound leg should be flown for 1 minute or 1 1/2 minutes (appropriate to altitude). Timing for subsequent outbound legs
should be adjusted, as necessary, to achieve proper inbound leg time.
Pilots may use any navigational means available; i.e., DME, RNAV,
etc., to ensure the appropriate inbound leg times.
(b) Outbound leg timing begins over/abeam the fix, whichever occurs
later. If the abeam position cannot be determined, start timing when
turn to outbound is completed.

In years past this was common for all holds, unless otherwise specified by the published hold instructions, or by ATC. All otherwise unspecified holds were assumed to be timed holds, including for GPS approaches. I only remember a few distance-defined holds in my local area from 10 years ago; now all those holds associated with GPS instrument approaches are distance defined.
Which brings us to the AIM 5-3-7 § j. 5. (emphasis on the last two sentences is mine):

5. Distance Measuring Equipment (DME)/ GPS Along-Track Distance (ATD). DME/GPS holding is subject to the same entry and holding procedures except that distances (nautical miles) are used in lieu of time values. The outbound course of the DME/GPS holding pattern is called the outbound leg of the pattern. The controller or the instrument approach procedure chart will specify the length of the outbound leg. The end of the outbound leg is determined by the DME or ATD readout. The holding fix on conventional procedures, or controller defined holding based on a conventional navigation aid with DME, is a specified course or radial and distances are from the DME station for both the inbound and outbound ends of the holding pattern. When flying published GPS overlay or stand alone procedures with distance specified, the holding fix will be a waypoint in the database and the end of the outbound leg will be determined by the ATD. Some GPS overlay and early stand alone procedures may have timing specified.

See also AIM Fig 5-3-7 and the associated note, which address the type of distance holds found on GPS approaches:

NOTE-
The inbound course is always toward the waypoint and the ATD is zero at the waypoint. The end of the outbound leg of the holding pattern is reached when the ATD reads the specified distance.

The holds found on GPS approaches will generally be distance based. The AIM states that some might still be time based, but I am no longer aware of any such holds in actuality. Every approach that I know of that used to have a time based hold on a GPS approach now has a distance based hold.
All holds should be flown as published or as instructed by ATC. If ATC does not instruct you otherwise, you are expected to fly the hold as published. In this case, that means you are expected to fly the hold as a distance-based hold. If you are not sure how you should fly the hold, ask ATC. They will provide you with the complete holding instructions if requested. See AIM 5-3-7 § b.:

b. If the holding pattern is charted and the controller doesn't issue
complete holding instructions, the pilot is expected to hold as
depicted on the appropriate chart. When the pattern is charted, the
controller may omit all holding instructions except the charted
holding direction and the statement AS PUBLISHED; e.g., HOLD EAST AS
PUBLISHED. Controllers shall always issue complete holding
instructions when pilots request them.

If the question is "do you have to fly the full distance indicated in the hold?", the answer is no, per a 2011 FAA legal interpretation. See below.

@fosterboondoggle That is not the question. The question is whether particular holds should be flown as timed OR distance legs. The AIM states that the pilot is expected to fly the hold as published. The FAA states, as you quote, that the pilot is not required to fly the full distance leg, i.e, will not be violated merely for omitting such an action.

J Walters - the reason for my comment was that if you search on google "do you have to fly the full distance indicated in the hold", this answer comes up with the top hit prominently excerpted as "short answer: yes". I understand that that may not be the intention of the original question, but that might be why it has 6000+ views, and I suspect that other readers (like myself originally) may be misled.

Yes. The number denotes the distance of the holding pattern. From FAA Instrument Flying Handbook:

DME and IFR-certified GPS equipment offer some additional options for holding. Rather than being based on time, the leg lengths for DME/GPS holding patterns are based on distances in nautical miles. These patterns use the same entry and holding procedures as conventional holding patterns.

[Modifying per Michael Hall's comments]
If the question is "do you have to fly the full distance indicated in the hold?", the answer is no. (But if it's just about timing vs. distance then ¯_(ツ)_/¯.)
I realize this is old, but since it's currently the top hit on google for the question "do you have to fly the distance shown on a depicted hold" I think it's worth revisiting. Google pulls out the accepted reply as "Short answer: yes" at the top of the search result. This appears to be incorrect. Another site led me to the "Young-HHC-2011" letter from the FAA, which among other things answers this question a bit differently from the accepted answer. Quoting from the letter:

... it is permissible, without specific ATC clearance, to
shorten published outbound DME legs in a holding pattern as long as the issued holding pattern leg length is not exceeded.

[emphasis mine]
It goes on to say that you have to comply with required leg lengths if ATC tells you to. (No surprise there.) But for those of us flying bug smashers at 90 kts on approach, this seems useful.

The OP is not asking whether it is permissible to shorten the leg, they are asking about the basis of the depicted hold.  I.e. whether it is based on distance, or whether timing is still expected, but distance is shown just for reference.  Your answer is useful, but the short answer to the base question should be "yes".

@MichaelHall are you sure? I read "Do we need to fly the leg as 4 nm, or the number conveys other meanings?" as "do we need to fly 4 nm"? The accepted answer is "the pilot is expected to fly the hold as depicted, with an outbound leg based on distance" which is not strictly correct even though that's what the AIM seems to say pretty literally. The interpretation lettter says you can fly less than the 4nm (or whatever - I saw a 7 nm HILPT the other day).

@MichaelHall, I also got to this by searching for answers on whether you have to fly the full distance depicted, and this post has been viewed a gazillion times. Maybe everyone's looking for answers to your version, but I somewhat suspect not since the accepted answer is right from the AIM.

I agree the question is poorly worded.  Perhaps I will chime more in later...

But yes, I am confident in my interpretation.  The title question specifically calls out timing vs distance, not shorter distance.

Btw, @FosterBoondoggle it may be appropriate to post a separate question titled "Do you have to fly the full distance indicated in the hold?" and post this answer there (you can answer your own question). I think it's clearly a question people are interesting in and would be be useful addition to the site.


text:
How to start an approach in IMC loss of communication under IFR?
Suppose I lost my radio communication while en-route, I am following the cleared route to the last fix. I am flying the higher of my last assigned altitude and MEA. Then, I am holding at my last en-route fix and the filed expected arrival time is coming. In this case, how may I start to descent and commerce an approach? regarding which IAP to use and which IAF to head for, provided that the last en-route fix is none of IAFs in any IAPs and my last altitude is much higher than the minimums prescribed in the IAPs.
POST EDIT:

Transponder is on 7600
No VMC within the range of fuel
Destination airport is tower-controlled

If you know that far in advance, I'd hope you're squawking 7600 by then. My guess is that ATC would then clear the way for you on your filed flight plan knowing they can't tell you to do anything else.

The FAA says (per 14 CFR 91.185):

If the clearance limit is not a fix from which an approach begins,
  leave the clearance limit at the expect-further-clearance time if one
  has been received, or if none has been received, upon arrival over the
  clearance limit, and proceed to a fix from which an approach begins
  and commence descent or descent and approach as close as possible to
  the estimated time of arrival as calculated from the filed or amended
  (with ATC) estimated time en route.

This seems to answer your question unless you have a more specific one. You can descend to an IAF as you see fit and commence any approach. ATC will clear the way.

So it is reasonable to make a VPATH from the last en-route fix to an IAF on approach plate?

@skyoasis You cannot start descent until your ETA, so if you are proceeding from your last enroute fix because no approaches start there, and you have not reached your ETA yet, then  you should not start descending yet, and should not use VPATH there.  Instead, you remain level until your ETA (while proceeding to your IAF) and only start your descent at your ETA, holding at the IAF if needed.

Let's extract some abbreviations from 14 CFR 91.185 (c) (3),

ECL: En-route clearance limit
EFC: Expected further clearance
IAF: Initial approach fix, a fix from which an approach
ETA: Estimated time of arrival, adjusted by departure time offset and superseded by most recent ATC amendment.

and expand the four possible concrete scenarios regarding how to commence   the approach to the destination airport.

ECL is a IAF with EFC:
En-Route → ECL/IAF (Hold until EFC) → Descent and approach
ECL is a IAF without EFC:
En-Route → ECL/IAF (Hold until ETA) → Descent and approach
ECL is not a IAF with EFC:
En-Route → ECL (Hold until EFC) → IAF (Hold until ETA) → Descent and approach
ECL is not a IAF without EFC:
En-Route → ECL → IAF (Hold until ETA) → Descent and approach

Comm failure questions are always interesting.
Flying at a significantly higher altitude because of a comm failure, and then having to get down, I would  chose an approach procedure that has a procedure turn or course reversal, and commence descent in a holding pattern over that procedure turn/reversal fix, at my ETA.
If there is no such approach, I would shuttle-descend over the chosen IAF on my inbound track to that IAF, making sure to go no lower than any minimum IFR altitude, and always remaining within the area of coverage of that minimum IFR altitude. I might tweak my inbound track a bit to avoid too sharp turns onto the intermediate or final approach course.
This could take quite some time if the flight plan was filed for FL410 and you suffer a total comm failure. While shuttling down from FL410, I might try to whip out a cellphone (and an ATC phone number).
Luckily comm failures, especially in the worst-case circumstances assumed here, don't happen too often.

What about the expected cellular signal strength and voice quality for cellphones at FL410?

@skyoasis I would not expect a cellular signal at FL410. But depending where you are, you might get a usable signal as high as FL200. I would only do this as an emergency measure, for example if low IMC prevailed everywhere along my route. In that case, I consider getting wx info to be absolutely critical. Again, this is all very hypothetical.

Also note that we have near universal radar over conus so they are probably watching you anyway.

@Ugo, 41,000 ft is not all that far. Just 12.5 km or 7¾ mi which should be well within range of GSM stations. The main problem with phone calls from aircraft is that the phone would be hopping from one tower to the next quickly, draining a lot of battery in the process. There may be actually more problem down low due to closure rate as that GSM has limit on that; 3G and 4G should be better off in this regard.


text:
Is it allowed to use an ex-military aircraft overseas?
We are considering using an ex US military helicopter in our counter poaching operations in the Central African Republic, while keeping it registered as a US aircraft with the US "N" number. We are not exporting it to sell and we will and we will maintain ownership and the US "N" number.

What restrictions are there for taking an aircraft like this out of the US, if any?

I know of a good many former US military helicopters in private and commercial uses in some African countries. And I know that some ex US military helicopters are not "eligible" to export, and some are.

What is the difference?

Before considering U.S. regulations, have you considered the African regulations?

Either today, or tomorrow, or day after, you're going to need a lawyer to do this. Why not start from there

Welcome to Aviation.SE! Please be aware that whatever answer you will get here, it does not constitute legal advice, you will still have to ask and deal with the appropriate authorities.

You might get an answer at this pilots forum, since aviation professionals who operate all over the world often visit and share their insights, but as @Federico points out, that's still advice pulled from the internet and the actual process to get something like this done takes a contract and overcoming a variety of legal hurdles.  Best of luck in your efforts.

I think you need to study the export regulation more than aviation regulation, e.g. if you export the aircraft with clear knowledge or intention to misuse or abuse it, then the action of export could get you in trouble. But the action of actually using it doesn't seem to be bounded by any US  regulations any more.

We have all of the Africa regs and permission form C.A.R. to operate an helicopter anywhere in that country.

From what I can understand Export means to sell it to another party which we are not doing.  We will maintain ownership.

@AfricanWildlifeInitiative That doesn't seem true in the context of US laws.  Again I recommend you consult a lawyer to verify that.

From my experience exporting and importing aircraft, IMO  @AfricanWildlifeInitiative is correct. Export in the context of the FAA is associated with a transfer of ownership. Operation is a totally different issue. But go talk to a lawyer who will stand behind his advice and interpretation of rules and regs.

If it has an N-number and an airworthiness certificate that allows the operations you want to perform with the aircraft, it doesn't matter what it did in a past life, you can operate wherever the country that controls the airspace will allow you to do so. As was mentioned before, the regulations of the country in which you wish to operate the aircraft are more relevant to what you want to do.
If on other other hand you have an ex-military aircraft that has an N-number but an exhibition-only experimental airworthiness certificate for a warbird, to pick an example, or some other type of restricted certificate, you're likely to be required to notify the FAA where and when you intend to fly it, and that may trigger additional oversight and possibly a denial of the permission to do so in a foreign country. The operating limitations document of the airworthiness certificate may also limit you to specific airspace within the US. Check the aircraft documentation and see if the FAA already put limits on the helo's operations.
That's where you will find the answer from the FAA point of view.


text:
What about hours flown solo without a student pilot certificate?
My instructor has ruined my training. 2.5 months ago when I started private pilot training (FAR61) I had zero knowledge about aviation.  After 1.5 months he signed me off for solo and I finished all the requirements and we sat together to register through IACRA for my checkride, and guess what? He had forgotten to tell me I needed a student pilot certificate before I go soloing.
Besides the risk he made on my life ( imagine I crashed during solo and insurance definitely would deny any claims since I was not legally allowed to fly! ) and that I will be grounded for weeks until I get student pilot certificate, I don’t know what will happen to those solo hours ?

Do I have to re-do the whole solo requirements or they will still count towards my hours?
Do I need to file a complaint through AOPA or FAA or contact my AOPA representative?
What are the rules and who is responsible for that?

Generally, the PIC is responsible for making sure all regulations are met, and is the final authority on the operation of the aircraft. (I don't know if that's the case in your specific situation, which is why I'm not posting this as an answer.) Sure, your instructor shouldn't have made such an omission, but I very strongly suspect that even a student PIC is expected to be familiar with the relevant regulations by the time of their first solo flight.

In order to pass the Knowledge Test you should have studied Part 61. Especially §61.3   Requirement for certificates, ratings, and authorizations. Required pilot certificate for operating a civil aircraft of the United States. No person may serve as a required pilot flight crewmember of a civil aircraft of the United States, unless that person:
(1) Has in the person's physical possession or readily accessible in the aircraft when exercising the privileges of that pilot certificate or authorization—
(i) A pilot certificate issued under this part and in accordance with §61.19;

FYI, You should immediately file a NASA form.

Your instructor didn't "ruin" your training, your hours still count (maybe not the solo ones). You must have gained enough knowledge so that he felt you were safe to fly. BTW, if you crashed, it wouldn't have only been your problem, it would have been his too, since the insurance company would likely go after him, not you. I'd find a new instructor, but you aren't totally lost (or grounded, you can still fly with an instructor!).

Just a little bit of friendly advice: I definitely feel that you should discuss this with your instructor, if nothing else then to let him know you're finding another and why, but do try to be a little less absolutist when you do than you come across in this question. "My instructor has ruined my training." As pointed out above, no, he most likely hasn't. You may feel like he has, and you're entitled to those feelings, but I encourage you to phrase it when talking to him in terms of your thoughts and feelings ("I fear/feel/worry/... that you have..."), rather than facts ("you have...").

What was the venue for your training? Who supplied the airplane?

Do I have to re-do the whole solo requirements or they will still count towards my hours?
You will probably not have to re-do the hours. An FAA examiner will look at your log, test you and decide whether to give you a license. It is unlikely that they would notice that your solo time was done before the student license was issued and then disallow the hours. If they did notice and disallowed the hours, you could appeal that decision.
As long as a CFI certified you for solo flight, the hours you did should count.
Do I need to file a complaint through AOPA or FAA or contact my AOPA representative?
No.
What are the rules and who is responsible for that?
It's your job as a pilot to know the regulations and follow them, including requirements such as obtaining student licenses and medicals. The CFI is not your nanny. It is YOUR responsibility to know the rules and follow them.

I agree that the student does carry responsibility, but so does the instructor, who should be in hot water over this. The CFI is not allowed to be sending students solo without verifying that all the boxes have been ticked. If that happened here in Australia CASA would take their CFI rating off them in an instant, he’d better hope the FAA is more lenient.

@Ben indeed. As should the aircraft owner who has responsibility to not allow a person without the correct papers to take control of an aircraft without an instructor on board.

@jwenting I'd be careful blaming the owner.  Many times, they lease their airplane to a school or an instructor and that gets delegated as a term of the lease.  In this situation, the owners often (usually?) don't even know who is flying their airplane, much less check them out on their own.

@Lnafziger in that case the responsibility goes from the owner to the person or company holding the lease, obviously.

@jwenting My point exactly....

The CFI MUST endorse your Student License for the particular model of aircraft PRIOR to your solo flying.  If you do not have the License, he can't endorse it, therefore you are not legal.
In some cases, through the years, CFIs have done the endorsement after you have completed your first solo. The FAA frowns on this.  The FAA would be incensed at the CFI in your case, and upset with you.
As far as allowing the hours logged, ask the FAA.

The Student License has changed (in the US) to where it is a permanent license and the license itself is not endorsed.  Instead the logbook is now endorsed for a student to solo.  In this case it sounds like the logbook was endorsed but the student never received the license.


text:
Why don't passenger planes move back by their own means before taxiing?
As far as I know, passenger planes needs a puller/pusher truck to go back in the land. But why they don't go back without truck assistance?

I’m not exactly sure, but it sounds like you’re asking why planes need a tug on pushback. Some jet and prop planes such as the 727 and MD-80 can perform a power back where they start at the gate and use reverse thrust to exit it. This procedure was more common in the 70s and 80s but seems to have died out in the 2000s. According to wikipedia the practice was stopped to save fuel. Additionally, there was a high risk of foreign object debris entering the engines from the ramps. Thus, the procedure was even less common on low engine aircraft such as the 737. The C-17 performs powerbacks still for the armed forces, so size is not a limiting factor.
You can check out a video here of an MD-80 powerback.


text:
Can I go into clouds if I got IFR clearance? (uncontrolled airport)
Just to make sure if I am right before I get in trouble.
If I am taking off from an uncontrolled airport for IFR flight, assuming I received IFR clearance on the ground, can I enter into the clouds before I contact approach (or center) after takeoff? Even when the Golf starts at the surface or 700 ft?
Would clearance give me a heading and altitude that I would want to follow right after takeoff?

What class of airspace are you in?

Which country, which airspace, and what exactly did the clearance say?

If you're unclear you should check with a CFII

Related -- https://aviation.stackexchange.com/questions/21888/can-an-ifr-clearance-be-issued-and-flown-through-imc-in-class-g-airspace?rq=1 , https://aviation.stackexchange.com/questions/79262/is-some-sort-of-clearance-such-as-a-clearance-to-enter-nearby-controlled-airspa?noredirect=1&lq=1, https://aviation.stackexchange.com/questions/38842/do-you-need-an-instrument-rating-to-fly-in-imc-in-class-g-airspace?noredirect=1&lq=1

This question is unclear-- "Even when the Golf starts at the surface or 700 ft?" -- Golf "starting" at 700 ft sounds like you mean 700 feet is the floor not the ceiling of Golf, which is confusing.  Especially in the context of the other possibility "Golf starts at the surface" which is obviously a reference to the floor of Golf not the ceiling.  Can you rephrase this sentence and clarify please?

For typical uncontrolled airports which have instrument approaches, they are usually in Class G airspace from the surface to 700’ AGL.  In this airspace, yes, you can - at least from a regulatory perspective fly in IMC without ATC clearance, provided you remain outside of controlled airspace.
In order to fly a cleared IFR flight plan, you would have to depart into uncontrolled airspace prior to climbing into the Class E shelf above the airport.  Typically, IFR departure clearances from uncontrolled airports contain a time window in which you must be airborne and in contact with a regional departure control or ARTCC by eg “Clearance void 30 minutes from now”.  Basically, in doing so, ATC have given you a block of airspace which you have use of for a small time window in order to get airborne and join ARTCC on your cleared flight plan.  Once you have departed the uncontrolled airport on CTAF, you would immediately go over to your assigned departure control frequency and report in.  The controllers would thence take over and direct your flight.
A typical IFR clearance from a untowered airport would be something like as follows:

Mooney Three One Two Seven Zulu, Atlanta Approach, cleared to the Charleston airport [or: Charlie Hotel Sierra airport] via direct NELLO, then as filed. Climb and maintain five thousand, expect one-one thousand five minutes after departure. Departure frequency one-two-one-point-zero, squawk two-four-seven-seven. Hold for release.

The "Hold for release" is telling you that you have a clearance to fly IFR on the given route, but not yet! You must have a release in order to legally depart the airport under Instrument Flight Rules (whether the weather is IMC or not) and be afforded IFR separation. You are still allowed to depart the airport under VFR (if the weather permits) but you should squawk 1200 if you do.
You will eventually hear...

Mooney Two Seven Zulu, released for departure. Clearance void if not off in three-zero minutes. If not off in three-zero minutes, advise Atlanta Approach of intentions within three-five minutes.

From the time the controller issues the release, you now have thirty minutes to get airborne, on a course direct to the NELLO fix and contact Atlanta Departure on 121.0 or the clearance becomes invalid and you must request a new clearance to do so.  If there are low clouds between you and the time you switch over to departure, so be it, you’re going into IMC.  Usually, though, it’s best once airborne and cleaned up, to immediately go over to departure control as quickly as you can.
Another example of IFR flight in Class G is flying an instrument approach into an uncontrolled airport.  Once you descend below the Class E shelf, the controller hands you off to CTAF and you continue on in uncontrolled airspace.  If you’re flying an ILS approach down to minimums, that may well be another 400-500 ft descending uncontrolled in IMC before you either break out and land or go missed, thence rejoin center to report a missed approach.

Similar to this question: https://aviation.stackexchange.com/questions/21888/can-an-ifr-clearance-be-issued-and-flown-through-imc-in-class-g-airspace?rq=1

You can get a clearance from an airport that has no instrument approaches or departures and is in Class G airspace. In that case, you are responsible for terrain clearance until you reach a point in controlled airspace where ATC can provide IFR services. If you are in the clouds before reaching controlled airspace, ATC will not be providing separation from other IFR aircraft, so you’d better hope no one else is up there.

Without an ATC clearance, you are under Visual Flight Rules (VFR). Therefore you may not enter clouds, as defined in the 14 CFR: https://www.gleim.com/aviation/faraim/?leafNum=91.155.

Related -- https://aviation.stackexchange.com/questions/21888/can-an-ifr-clearance-be-issued-and-flown-through-imc-in-class-g-airspace?rq=1 , https://aviation.stackexchange.com/questions/79262/is-some-sort-of-clearance-such-as-a-clearance-to-enter-nearby-controlled-airspa?noredirect=1&lq=1, https://aviation.stackexchange.com/questions/38842/do-you-need-an-instrument-rating-to-fly-in-imc-in-class-g-airspace?noredirect=1&lq=1

@Mark Jones Jr. -- re "Without an ATC clearance, you are under Visual Flight Rules (VFR)." -- can you prove this statement to be true or at least explain in more detail why you think it to be true?  This related answer https://aviation.stackexchange.com/a/92543/34686 by a person who is or has been an ATC controller (see for example https://aviation.stackexchange.com/a/83970/34686), says you don't need an IFR clearance to legally fly IFR in IMC in uncontrolled airspace.

@quietflyer "91.123 Compliance with ATC clearances and instructions.
(a) When an ATC clearance has been obtained, no pilot in command may deviate from that clearance unless an amended clearance is obtained, an emergency exists, or the deviation is in response to a traffic alert and collision avoidance system resolution advisory. However, except in Class A airspace, a pilot may cancel an IFR flight plan [cancelling an ATC clearance] if the operation is being conducted in VFR weather conditions."

@MarkJonesJr. -- I don't follow how that addresses my comment.  If you don't have an ATC clearance in the first place, the issue of whether you are deviating from a clearance is irrelevant.

Yes.  First note that takeoff from any IFR qualified airport will allow for takeoff and avoidance of obstacles by traditional TERPS methods.  Therefore, the type of airspace has no effect on IFR obstacle avoidance.
 Thus, you can enter the weather and be assured of clearance from obstacles.  However, you will not be provided separation from VFR aircraft in the uncontrolled airspace that is free of weather.  You will get separation from IFR aircraft. You may not get a heading, especially if you are conducting a diverse departure, but your IFR clearance will have an altitude.

I usually had a heading assigned to enter controlled airspace.  IE -  Enter controlled airspace on a heading of 180.


text:
Can I get pop-up IFR clearance to waypoint (intersection)?
Let's say I departed from the airport to fly to the practice area (VFR).
However, there is a thick cloud layer between where I am at right now and practice area. I already got a report that practice area is clear and VFR.
To go through the cloud, I need to request IFR to go through the cloud.
At this point, can I request pop-up IFR to any waypoint (as a destination) that is in practice area?
Will I be able to cancel that IFR flight right after I go through the cloud and when I am not in IMC?

Welcome to aviation.SE! Which country or regulations are you asking about?

Yes, you can file to a waypoint or navaid and can cancel IFR anytime you want. Whether or not ATC will grant a popup clearance depends on the facility and their workload. In places with a lot of flight training (Florida, Southern California), ATC, in particular the TRACONs, usually grants popups.


text:
Is skydiving in IMC (through clouds) legal and safe?
(1) What are the legal considerations for skydiving into clouds? I see many skydivers did fall into and out clouds in online videos. However, the clouds are mostly scattered or thin-layered. How about in widespread hard-IMC, say OVC 008?
(2) What about maintaining the actual safety? How does a skydiver maintain attitude and stable position in IMC and not disoriented? and separation from other traffic?
EDIT: The legal consideration of my context is within US, but open to legal regulations worldwide.

The part belonging to Aviation is about the conditions required for a pilot to legally release skydivers. The rest is off-topic. IMC may not be the criteria taken into consideration, and this may vary by country, You may consider adjusting your post.

A skydiver shares airspace with the aviation world. It is worth knowing the responsibilities of both the pilots of the jump airplanes and the skydivers.

In scandinavia, we do jump through clouds, as long as the base is high enough for clear sight, at separation (still freefall), for deployment of parachute, and canopy flight. cloud base in 4000 feet is the normal minimum, although you can actually exit the plane from lower altitudes as well. All jumpers are required to have a flying canopy at 2000 feet at least, meaning you should know your canopy's opening time/distance. I normally deploy at 3000 feet, and have a fully functional parachute between 2500 and 2000

It is not legal to conduct skydiving into clouds, atleast in US. From 14 CFR §105.17   Flight visibility and clearance from cloud requirements:

No person may conduct a parachute operation, and no pilot in command of an aircraft may allow a parachute operation to be conducted from that aircraft—
(a) Into or through a cloud, or
(b) When the flight visibility or the distance from any cloud is less than that prescribed in the following table:

Is it safe? Probably. The chances of mid-air collision are pretty small, but they can happen as is does with aircraft. Midair collisions between skydivers and between aircraft and skydivers are rare, but not unheard of. There has also been reports of near misses between aircraft and skydivers through clouds. So, its better to be safe.

http://www.nytimes.com/1993/11/23/us/4-die-after-their-plane-collides-with-a-sky-diver.html

The BBC video isn't available to US viewers.

Is it legal? No. Skydiving is a VFR operation, FAA regulations ( assuming the USA ) clearly say no jumping through clouds. Few jump planes are IFR equipped to get up there anyway.
Is it safe? Yes. Done it plenty of times myself on partly overcast days. The chances of hitting another aircraft in freefall are miniscule, and skydivers basically have no chance to see and avoid an aircraft anyway - the speeds and angles are too high. The pilot announces what's happening to other traffic and it's over in a minute. Determining the exit point with >50% cloud cover is not easy but GPS has largely eliminated that issue.
Flying a parachute through cottonball-cumulus clouds happens, it's kinda boring. Going around tall formations is much more interesting but doesn't really qualify as "into". If the weather systems are strong enough to affect a parachute's handling you won't find any canopies up there as everyone is in the local bar.

Downvote because of, 1) the statements which amount to admission and condoning of action in violation of safety-based aviation regulations, and 2) statements that such actions are safe despite listing some of the very real risks associated with such violations, however miniscule such risks may or may not be. This post provides a good case study in some of the hazardous attitudes in aviation.

The capability of a jump plane to climb up through a cloud, is only peripherally related to its capability to release a jumper above a cloud.

I think you're substituting "safe" for unlikely.